n2 + 3h2 -> 2nh3\nhow many moles of n2 are needed to produce 7 moles of nh3?\nround to keep 2 decimal…

n2 + 3h2 -> 2nh3\nhow many moles of n2 are needed to produce 7 moles of nh3?\nround to keep 2 decimal places. ex:( 5.00moleh2o)\n4 nh3 prod requires n2 balanced eq. n2 + 3 h2 -> 2 nh3\n1 mole n2 = 2 mole nh3\n2 given 7.00 mole nh3\n3 want mole n2\n4 set - up mole n2 = 7×(1÷2)\n5 3.50 mole n2

n2 + 3h2 -> 2nh3\nhow many moles of n2 are needed to produce 7 moles of nh3?\nround to keep 2 decimal places. ex:( 5.00moleh2o)\n4 nh3 prod requires n2 balanced eq. n2 + 3 h2 -> 2 nh3\n1 mole n2 = 2 mole nh3\n2 given 7.00 mole nh3\n3 want mole n2\n4 set - up mole n2 = 7×(1÷2)\n5 3.50 mole n2

Answer

Explanation:

Step1: Identify mole - ratio

From the balanced equation $N_2 + 3H_2\rightarrow2NH_3$, the mole - ratio of $N_2$ to $NH_3$ is $1:2$, i.e., 1 mole of $N_2$ produces 2 moles of $NH_3$.

Step2: Set up proportion

Let $x$ be the number of moles of $N_2$ needed to produce 7 moles of $NH_3$. Using the mole - ratio, we have the proportion $\frac{x}{7}=\frac{1}{2}$.

Step3: Solve for $x$

Cross - multiply to get $x = 7\times\frac{1}{2}=3.50$.

Answer:

3.50 mole N2