10. how many grams of h₂ is produced by the reaction of 200 ml of a $\frac{0.88 \text{ moles}}{l}$ h₃(po₄)…

Explanation:

Step1: Calculate moles of $H_3(PO_4)$

Use the formula $n = cV$, where $c$ is concentration and $V$ is volume. Given $c = 0.88\ mol/L$ and $V=200\ mL = 0.2\ L$. So $n(H_3(PO_4))=0.88\ mol/L\times0.2\ L = 0.176\ mol$.

Step2: Calculate mass of $H_3(PO_4)$

The molar - mass of $H_3(PO_4)$ is $M=(3\times1 + 31+4\times16)\ g/mol=98\ g/mol$. Using the formula $m = nM$, we get $m(H_3(PO_4))=0.176\ mol\times98\ g/mol = 17.248\ g$.

Step3: Determine mole - ratio from the balanced equation

From the equation $2Cr + 2H_3(PO_4)\rightarrow3H_2 + 2Cr(PO_4)$, the mole - ratio of $H_3(PO_4)$ to $H_2$ is $2:3$.

Step4: Calculate moles of $H_2$

If $n(H_3(PO_4)) = 0.176\ mol$, then $n(H_2)=\frac{3}{2}\times n(H_3(PO_4))=\frac{3}{2}\times0.176\ mol = 0.264\ mol$.

Step5: Calculate mass of $H_2$

The molar - mass of $H_2$ is $M = 2\ g/mol$. Using the formula $m = nM$, we get $m(H_2)=0.264\ mol\times2\ g/mol = 0.528\ g$.

Answer:

$0.528\ g$